Mathematics · 2022
JEE Main · 28 June 2022, Shift 2 · Q10
Let x=x(y) be the solution of the differential equation 2 y e^x / y^2 d x+(y^2-4 x e^x / y^2) d y=0 such that x(1)=0. Then, x( e ) is equal to:
Let $\displaystyle x=x(y)$ be the solution of the differential equation $\displaystyle 2 y \mathrm{e}^{x / y^2} \mathrm{~d} x+\left(y^2-4 x \mathrm{e}^{x / y^2}\right) \mathrm{d} y=0$ such that $\displaystyle x(1)=0$. Then, $\displaystyle x(\mathrm{e})$ is equal to :
Official answer
(4)
$\displaystyle -\mathrm{e}^2 \log _{\mathrm{e}}(2)$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.